<HTML>

<HEAD>

<TITLE> Naturally Accelerated Motion</TITLE>

<META NAME="GENERATOR" CONTENT="Internet Assistant for Microsoft Word 2.0z">
</HEAD>

<BODY bgcolor="#ffffff">

<p><CENTER><IMG src="banner_nat_accel_motion.jpg" ALT="Naturally Accelerated Motion"><p>

<I>Michael Fowler</I>
<br>
<I>UVa Physics Department</I></center>
<p>
<A HREF="lecturelist.html">Index of Lectures and Overview of the Course</A><BR>
<A HREF="gal_accn96.htm">Link to Previous Lecture</A><P> 

<H3>Distance Covered in Uniform Acceleration </H3>

<P>
In the last lecture, we stated what we called Galileo's acceleration
hypothesis:
<P>
<I>A falling body accelerates uniformly: it picks up equal amounts
of speed in equal time intervals, so that, if it falls from rest,
it is moving twice as fast after two seconds as it was moving
after one second, and moving three times as fast after three seconds
as it was after one second.</I>
<P>
We also found, from the experiment, that a falling body will fall
four times as far in twice the time. That is to say, we found
that the time to roll one-quarter of the way down the ramp was
one-half the time to roll all the way down.
<P>
Galileo asserted that the result of the rolling-down-the-ramp
experiment confirmed his claim that the acceleration was uniform.
Let us now try to understand why this is so. The simplest way
to do this is to put in some numbers. Let us assume, for argument's
sake, that the ramp is at a convenient slope such that, after
rolling down it for one second, the ball is moving at four feet
per second. This means that after two seconds it would be moving
at eight feet per second, after three seconds at twelve feet per
second and so on until it hits the end of the ramp. 
<P>
<I>To get a clear idea of what's happening, <B>you should sketch
a graph of how speed increases with time</B>. This is a straight line
graph, beginning at zero speed at zero time, then going through
a point corresponding to four feet per second at time one second,
eight at two seconds and so on. It sounds trivial, but is surprisingly
helpful to have this graph in front of you as you read -- so, find a 
piece of paper or an old envelope (this doesn't have to be too precise) 
and draw a line along the bottom marked 0, 1, 2, for seconds of time, 
then a vertical line (or y-axis) indicating speed at a given time -- 
this could be marked 0, 10, 20,...feet per second. Now, put in the 
points (0,0), (1,4) and so on, and join them with a line.</I>
<P>
From your graph, you can now read off its speed not just at 0, 1, 2 
seconds, but at, say 1.5 seconds or 1.9 seconds or any other time 
within the time interval covered by the graph.
<P>
The hard part, though, is figuring out how <I>far</I> it moves in a given
time. This is the core of Galileo's argument, and it is essential that 
you understand it before going further, so <I><B>read the next paragraphs 
slowly and carefully!</I></B>
<P>  
<I>Let us ask a specific question:</I> how far does it get in two
seconds? If it were moving <I>at a steady speed</I> of eight feet
per second for two seconds, it would of course move sixteen feet.
But it can't have gotten that far after two seconds, because it
just attained the speed of eight feet per second when the time
reached two seconds, so it was going at slower speeds up to that
point. In fact, at the very beginning, it was moving very slowly.
Clearly, <I>to figure out how far it travels during that first two
seconds what we must do is to find its <B>average</B> speed during
that period. 
<P>
This is where the assumption of <B>uniform </B>acceleration
comes in</I>. What it means is that the speed starts from zero at
the beginning of the period, <I>increases at a constant rate</I>, is four feet
per second after one second (half way through the period) and
eight feet per second after two seconds, that is, at the end of
the period we are considering. Notice that the speed is two feet
per second after half a second, and six feet per second after
one-and-a-half seconds. From the graph you should have drawn above
of the speed as it varies in time, it should be evident that, for this 
<I>uniformly</I> accelerated motion, the <I>average speed</I> over this two second
interval <I>is the speed reached at half-time</I>, that is, four
feet per second. 
<P>
Now, the distance covered in any time interval is equal to the 
average speed multiplied by the time taken, so the distance traveled 
in two seconds is eight feet---that is, four feet per second for two seconds.
<P>
<I>Now let us use the same argument to figure how far the ball rolls
in just one second.</I> At the end of one second, it is moving at
four feet per second. At the beginning of the second, it was at
rest. At the half-second point, the ball was moving at two feet
per second. By the same arguments as used above, then, the average
speed during the first second was two feet per second. Therefore,
<I>the total distance rolled during the first second is just two
feet.</I>
<P>
We can see from the above why, in uniform acceleration, <I>the ball
rolls <B>four times</B> as far when the time interval <B>doubles</B></I>. If the
average speed were the <I>same</I> for the two second period as for the
one second period, the distance covered would <I>double</I> since the
ball rolls for twice as long a time. But since the speed is steadily
increasing, <I>the average speed also doubles.</I> It is the <I>combination</I>
of these two factors---moving at twice the average speed for twice
the time---that gives the factor of four in distance covered!

<P>
It is not too difficult to show using these same arguments that
the distances covered in 1, 2, 3, 4, ...seconds are proportional
to 1, 4, 9, 16, .., that is, the squares of the times involved.
This is left as an exercise for the reader.
<H3>A Video Test of Galileo's Hypothesis </H3>

<P>
In fact, using a video camera, we can check the hypothesis of
uniform acceleration very directly on a falling object. We drop
the ball beside a meter stick with black and white stripes each
ten centimeters wide, so that on viewing the movie frame by frame,
we can estimate where the ball is at each frame. Furthermore,
the camera has a built-in clock---it films at thirty frames per
second. Therefore, we can constantly monitor the speed by measuring
how many centimeters the ball drops from one frame to the next.
Since this measures distance traveled in one-thirtieth of a second,
we must multiply the distance dropped between frames by thirty
to get the (average) speed in that short time interval in centimeters
per second. 
<P>
By systematically going through all the frames showing the ball
falling, and finding the (average) speed for each time interval,
we were able to draw a graph of speed against time. It was a little
rough, a result of our crude measuring of distance, but it was
clear that speed was increasing with time at a steady rate, and
in fact we could measure the rate by finding the speed reached
after, say half a second. We found that, approximately, the rate
of increase of speed was ten meters (1,000 cms) per second in
each second of fall, so after half a second it was moving at about
five meters per second, and after a quarter of a second it was
going two and a half meters per second. 
<P>
This rate of increase of speed is the same for all falling bodies,
neglecting the effect of air resistance (and buoyancy for extremely
light bodies such as balloons). It is called the <I>acceleration
due to gravity</I>, written <I>g</I>, and is actually close to
9.8 meters per second per second. However, we shall take it to
be 10 for convenience. 
<H3>Throwing a Ball Upwards</H3>

<P>
To clarify ideas on the acceleration due to gravity, it is worth
thinking about throwing a ball vertically upwards. If we made
another movie, we would find that the motion going upwards is
like a mirror image of that on the way down-the distances traveled
between frames on the way up get shorter and shorter. In fact,
the ball on its way up <I>loses</I> speed at a steady rate, and
the rate turns out to be ten meters per second per second-the
same as the rate of increase on the way down. For example, if
we throw the ball straight upwards at 20 meters per second (about
40 mph) after one second it will have slowed to 10 meters per
second, and after two seconds it will be at rest momentarily before
beginning to come down. After a total of four seconds, it will
be back where it started. 
<P>
An obvious question so: how high did it go? The way to approach
this is to find its average speed on the way up and multiply it
by the time taken to get up. As before, it is helpful to sketch
a graph of how the speed is varying with time. The speed at the
initial time is 20 meters per second, at one second it's down
to 10, then at two seconds it's zero. It is clear from the graph
that the average speed on the way up is 10 meters per second,
and since it takes two seconds to get up, the total distance traveled
must be 20 meters.
<H3>Speed and Velocity</H3>

<P>
Let us now try to extend our speed plot to keep a record of the
entire fall. The speed drops to zero when the ball reaches the
top, then begins to increase again. We could represent this by
a V-shaped curve, but it turns out to be more natural to introduce
the idea of velocity. Unfortunately velocity and speed mean the
same thing in ordinary usage, but in science velocity means more:
it includes speed and direction. In the case of a ball going straight
up and down, we include direction by saying that motion upwards
has positive velocity, motion downwards has negative velocity.

<P>
If we now plot the velocity of the ball at successive times, it
is +20 initially, +10 after one second, 0 after two seconds, -10
after three seconds, -20 after four seconds. If you plot this
on a graph you will see that it is all on the same straight line.
Over each one-second interval, the velocity decreases by ten meters
per second throughout the flight. In other words, the acceleration
due to gravity is -10 meters per second per second, or you could
say it is 10 meters per second per second <I>downwards</I>. 
<H3>What's the Acceleration at the Topmost Point?</H3>

<P>
Most people on being asked that for the first time say zero. That's
wrong. But to see why takes some clear thinking about just what
is meant by velocity and acceleration. Recall Zeno claimed motion
was impossible because at each instant of time an object has to
be in a particular position, and since an interval of time is
made up of instants, it could never move. The catch is that a
second of time cannot be built up of instants. It can, however,
be built up of <I>intervals</I> of time each as short as you wish.
Average velocity over an interval of time is defined by dividing
the distance moved in that interval by the time taken---the length
of the interval. We define <I>velocity at an instant of time</I>,
such as the velocity of the ball when the time is one second,
by taking a small time interval which includes the time one second,
finding the average velocity over that time interval, then repeating
the process with smaller and smaller time intervals to home in
on the answer.   
<P>
Now, to find acceleration at an instant of time we have to go
through the same process. Remember, acceleration is rate of change
of velocity. So, to find the acceleration at an instant we have
to take some short but non-zero time interval that includes the
point in question and find how much the velocity changes during
that time interval. Then we divide that velocity change by the
time it took to find the acceleration, in, say, meters per second
per second.
<P>
The point is that at the topmost point of the throw, the ball
does come to rest for an instant. Before and after that instant,
there is a brief period where the velocity is so small it looks
as if the ball is at rest. Also, our eyes tend to lock on the
ball, so there is an illusion that the ball has zero velocity
for a short but non-zero period of time. But this isn't the case.
The ball's velocity is always changing. To find its acceleration
at the topmost point, we have to find how its velocity changes
in a short time interval which includes that point. If we took,
for example, a period of one-thousandth of a second, we would
find the velocity to have changed by one centimeter per second.
So the ball would fall one two-thousandth of a centimeter during
that first thousandth of a second from rest-not too easy to see!
 The bottom line, though, is that the acceleration of the ball
is 10 meters per second per second downwards throughout the flight.

<P>
If you still find yourself thinking it's got no acceleration at
the top, maybe you're confusing velocity with acceleration. All
these words are used rather loosely in everyday life, but we are
forced to give them precise meanings to discuss motion unambiguously.
In fact, lack of clarity of definitions like this delayed understanding
of these things for centuries.
<H3>The Motion of Projectiles </H3>

<P>
We follow fairly closely here the discussion of Galileo in Two
New Sciences, <A HREF="http://galileoandeinstein.phys.virginia.edu/tns244.htm">Fourth Day</A>,
 from page 244 to the middle of page 257. 
<P>
To analyze how projectiles move, Galileo describes two basic types
of motion:
<P>
(i) Naturally accelerated vertical motion, which is the motion
of a vertically falling body that we have already discussed in
detail.
<P>
(ii) Uniform horizontal motion, which he defines as straight-line
horizontal motion which covers equal distances in equal times.
<P>
This uniform horizontal motion, then, is just the familiar one
of an automobile going at a steady speed on a straight freeway.
Galileo puts it as follows:
<P>
<I>&quot;Imagine any particle projected along a horizontal plane
without friction; then we know...that this particle will move
along this same plane with a motion that is uniform and perpetual,
provided the plane has no limits.&quot;</I>
<P>
This simple statement is in itself a substantial advance on Aristotle,
who thought that an inanimate object could only continue to move
as long as it was being pushed. Galileo realized the crucial role
played by friction: if there is no friction, he asserted, the
motion will continue indefinitely. Aristotle's problem in this
was that he observed friction-dominated systems, like oxcarts,
where motion stopped almost immediately when the ox stopped pulling.
Recall that Galileo, in the rolling a ball down a ramp experiment,
went to great pains to get the ramp very smooth, the ball very
round, hard and polished. He knew that only in this way could
he get reliable, reproducible results. At the same time, it must
have been evident to him that if the ramp were to be laid flat,
the ball would roll from one end to the other, after an initial
push, with very little loss of speed.
<H3>Compound Motion </H3>

<P>
Galileo introduces projectile motion by imagining that a ball,
rolling in uniform horizontal motion across a smooth tabletop,
flies off the edge of the table. He asserts that when this happens,
the particle's horizontal motion will continue at the same uniform
rate, but, in addition, it will acquire a downward vertical motion
identical to that of any falling body. He refers to this as a
<I>compound</I> motion.
<P>
<IMG SRC="projecgl.gif" ALIGN="BOTTOM">
<P>
The simplest way to see what is going on is to study Galileo's
diagram on page 249, which we reproduce here.
<P>
Imagine the ball to have been rolling across a tabletop moving
to the left, passing the point <I>a</I> and then going off the
edge at the point<I> b</I>. Galileo's figure shows its subsequent
position at three equal time intervals, say, 0.1 seconds, 0.2
seconds and 0.3 seconds after leaving the table, when it will
be at <I>i</I>, <I>f</I>, and <I>h</I> respectively. 
<P>
The first point to notice is that the horizontal distance it has
travelled from the table increases uniformly with time: <I>bd</I>
is just twice <I>bc</I>, and so on. That is to say, its horizontal
motion is just the same as if it had stayed on the table. 
<P>
The second point is that its vertical motion is identical to that
of a vertically falling body. In other words, if another ball
had been dropped vertically from <I>b</I> at the instant that
our ball flew off the edge there, they would always be at the
same vertical height, so after 0.1 seconds when the first ball
reaches <I>i</I>, the dropped ball has fallen to <I>o</I>, and
so on. It also follows, since we know the falling body falls four
times as far if the time is doubled, that <I>bg</I> is four times
<I>bo</I>, so for the projectile <I>fd</I> is four times <I>ic.</I>
This can be stated in a slightly different way, which is the way
Galileo formulated it to prove the curve was a parabola:
<P>
The ratio of the vertical distances dropped in two different times,
for example <I>bg</I>/<I>bo</I>, is always the square of the ratio
of the horizontal distances travelled in those times, in this
case <I>fg</I>/<I>io</I>.
<P>
You can easily check that this is always true, from the rule of
uniform acceleration of a falling body. For example,<I> bl</I>
is nine times <I>bo</I>, and <I>hl</I> is three times <I>io</I>.
<P>
Galileo proved, with a virtuoso display of Greek geometry, that
the fact that the vertical drop was proportional to the square
of the horizontal distance meant that the trajectory was a parabola.
His definition of a parabola, the classic Greek definition, was
that it was the intersection of a cone with a plane parallel to
one side of the cone. Starting from <I>this</I> definition of
a parabola, it takes quite a lot of work to establish that the
trajectory is parabolic. However, if we define a parabola as a
curve of the form y =Cx&#178; then of course we've proved it already!

<P>
Written material Copyright &copy; Michael Fowler 1996 except where
otherwise noted.
<P><A HREF="vectors.htm">Link to Next Lecture</A><P> 
<A HREF="lecturelist.html">Index of Lectures and Overview of the Course</A><BR>
<P>
</BODY>

</HTML>

